Let $$f\left( x \right) = \int\limits_1^x {\sqrt {2 - {t^2}} \,dt.} $$ Then the real roots of the equation
$${x^2} - f'\left( x \right) = 0$$ are

Let $$T>0$$ be a fixed real number . Suppose $$f$$ is a continuous
function such that for all $$x \in R$$, $$f\left( {x + T} \right) = f\left( x \right)$$.

If $$I = \int\limits_0^T {f\left( x \right)dx} $$ then the value of $$\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $$ is

Let $$T>0$$ be a fixed real number . Suppose $$f$$ is a continuous
function such that for all $$x \in R$$, $$f\left( {x + T} \right) = f\left( x \right)$$.

If $$I = \int\limits_0^T {f\left( x \right)dx} $$ then the value of $$\int\limits_3^{3 + 3T} {f\left( {2x} \right)dx} $$ is

If $${\overrightarrow a }$$ and $${\overrightarrow b }$$ are two unit vectors such that $${\overrightarrow a + 2\overrightarrow b }$$ and $${5\overrightarrow a - 4\overrightarrow b }$$ are perpendicular to each other then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is